Sunday, October 19, 2014

Q: The following challenge is based on a puzzle from a Martin Gardner book, that may not be well-known. Out of a regular grade school classroom, two students are chosen at random. Both happen to have blue eyes. If the odds are exactly 50-50 that two randomly chosen students in the class will have blue eyes: How many students are in the class?

It's going to be hard to provide hints to the answer this week.

Edit: My hint was "Be" which is the symbol for Berylium, element 4 on the periodic table. While 4 is a technically correct answer (3 blue-eyed children --> 3/4 x 2/3 = 1/2) the intended answer was for a more typical class size.

A: 21 students (15 with blue eyes and 7 without).

The probability the first child has blue eyes is 15/21 or 5/7. Once that child is taken out of consideration, there are 14 children with blue eyes out of 20 so the probability is 7/10. Multiplying that together, we have 5/7 x 7/10 = 5/10 = 1/2

153 comments:

Here's my standard reminder... don't post the answer or any hints that could lead directly to the answer (e.g. via a chain of thought, or an internet search) before the deadline of Thursday at 3pm ET. If you know the answer, click the link and submit it to NPR, but don't give it away here.

You may provide indirect hints to the answer to show you know it, but make sure they don't give the answer away. You can openly discuss your hints and the answer after the Thursday deadline. Thank you.

I had the privilege of working with Kate Bosworth on Blue Crush; we had to replace the face of the professional surfer with her face. She has the most striking eyes, one blue, one half-blue half-hazel; I might have looked a little longer than absolutely necessary. And of course, when the director gave me the squirt bottle to wet down her t-shirt for a shot; I just handed it to her and let her do it as much as she felt appropriate :)

Once saw a dog with one blue and one brown eye. Getting back to a careful reading of the challenge: "regular grade school classroom," & regardless of all the eye color nonsense, "how many students in this regular grade school classroom/"

I have a cat with one blue and one green eye. According to my spouse (who works in hearing research) that means he is deaf in the ear on the blue side. We have not tested this, but know he can hear at least on one side.

When I first heard and thought about the new puzzle, while still in bed, I thought it might be difficult and complicated, requiring math skills I do not possess, but then I began thinking about it logically and believe I have the only possible answer unless one of the kids is a Cyclops. Also I believe there is a bit of a trick in the presentation.

Assuming I've figured the probability correctly, and haven't fallen for some "trick" in the wording of the puzzle, I have one "regular" answer, two rather "irregular" answers, and no proof that those are the only three answers. Any other answers, however, would be even more "irregular" than those I found.

I don't think it gives anything away to observe that if there are B blue-eyed students out of a total of T students in the class, then the probability of the choosing one blue-eyed student at random is B/T, and the probability of choosing another blue-eyed student at random is (B-1)/(T-1), so the probability of choosing 2 blue-eyed students is (B/T)*(B-1)/(T-1). The puzzle is asking for B and T such that this expression equals 1/2.

ron was criticizing jan's observation and I was just pointing out that jan has it exactly right. Also, two source-posts below you see my observation essentially declaring that there infinitely many valid sets of B and T. What's to overthink?

Interesting point, WW! Recently I have been wondering what a BLT w/onion would taste like. But, not so long ago, I was in a restaurant and ordered a BLT, and forgot all about the onion. Next time, would you be good enough to remind me?

The L need be between them only orthographically. Frankly, I don't care what in order the B, L, and T appear, as long as it's mine. The three of them need to be between slices of rye toast with lots of mayo, which begs the question, why some people feel the need to add A? If you've got the mayo, who needs A, and if you don't, what's the point, he asked delicately?

There's one answer that's so trivial that I wonder if they'll even accept it. I submitted that, as well as what I'm sure is the intended answer, and then went on to say that if there's no limit to the class size, then there's no limit to the number of solutions!

In other words, for a complete graph on n vertices, find an n > 2 with a coloring of its vertices (blue and non-blue) such that the total number of edges connecting two blue vertices equals n(n – 1)/4 ∈ ℕ.

I think jan and I have already given the ultimate hints; but I'll offer one more:

Just as there are infinitely many numbers which are squares, but are also triangular numbers (i.e. 1, 36, 1225, etc.), and just as there are infinitely many Pythagorean triples in which the first two numbers differ by just one (i.e. 3, 4, and 5; 20, 21, and 29; 119, 120, and 169; etc.), so there are also infinitely many class-sizes with associated lesser numbers of blue-eyed students within such classes so that randomly picking two of them gives exactly even odds of both having blue eyes.

After a little playing around in a spreadsheet, I found the appropriate sequence on OEIS that gives the infinite family of solutions, but for reasonable class sizes (say, less than 500 students) there are only three answers.

After my bike ride today I eventually ended up at a wine tasting at a nearby supermarket where I shop. The rep. attending with the in store woman who does a rather pour job, by the way, and I got into a discussion of today's puzzle. I told him that it was reminiscent of the old puzzle where you wake up before dawn and don't want to turn on a light in the bedroom in order not to awaken your wife. You have a drawer with 20 pairs of black socks and also 20 pairs of gray socks, but they are not paired. In order to be sure you remove a matching pair of socks, no matter which color, how many socks is the minimum you must take? He thought and then figured it would be just over half. I said it was only three socks. No matter how many socks are in the drawer it is always 3. The number of socks given in the puzzle is a distraction and in no way important to solving the problem. As to this puzzle, I would suggest anyone having problems solving it consider it in context of the upcoming celebration of Halloween. Imagine you have a carved out, medium size, pumpkin with both black and orange jelly beans inside. Continue asking he puzzle in that context. WARNING! DO NOT FORGET TO FACTOR IN THE GENE CHROMOSOMES. I kid you not.

I''m not convinced there are an infinite number of solutions. It seems to me that B/T has to approach sqrt(2)/2 as T get infinitely large. That's not going to yield a lot of integer solutions. (I hope you're not planning to cut those cute little blue-eyed student into pieces!)

A curious result: For class sizes in the solution set, the total number of students divided by the number of students with blue eyes approaches the square root of two as the total number of students increases. Can anyone explain why this is?

PC: See Jan's notation and equation posted at 11:36 am, For large values of T and B, B(B-1) is close to B^2 and T(T-1) is close to T^2.On Thursday, I'll be interested to see the demonstration showing an infinite number of integer solutions.

In my junior of year of college, I took a full year (three quarters) of probability and statistics. Probability was, in fact, a minor portion of the course. In the first couple of weeks of the first quarter, I learned to solve problems like this one. Most of the course was theoretical statistics, not the “cookbook” statistics that was a prerequisite for Introduction to Psychology, which I took with my spouse. (We got married between sophomore and junior year.) That course, along with a full year of numerical analysis led me to the realization that I did not want to become a theoretical mathematician.

Out of a regular collegiate-grade school psychology classroom, two students are chosen at random, David and his spouse. Both happen to have blue eyes. The vision of the two randomly chosen students in the class is exactly 20-20 One of them is chosen to survey and record the eye color of all students in the classroom, which includes celebrities Paul Newman, Robert Redford, Frank Sinatra and John Wayne who was in the midst of filming “True Grit.” There are no mirrors in the classroom, just smoke. (It’s the 1970s!)

What is the minimum number of blue eyes that will be recorded? (And no, not recorded by Crystal Gayle Sayers, who played for the Bears, not the Browns! No, recorded rather, with great precision, by David or his spouse.)

This should be a "show your work" question. One could arrive at the answer through guessing or even doing it wrong. Obtaining the correct answer by the correct method is really the essence of this challenge.

LMP:BRAVO! I have been planning on making a similar comment on Thursday. It would be most interesting to learn what percentage of submissions are falsely arrived at, and therefor meaningless. Also I agree with Chuck that this is not really a math puzzle, but rather it is a logic puzzle and far more satisfying if solved that way.

I must admit that my initial efforts to solve this puzzle led me to some erroneous sequences of steps. Then I realized that this was NPR, and put the right combination of facts and logic together to arrive at what I believe is the correct solution,

An erroneous sequences of steps (a permutation--more on this later) will get you an answer that is partially correct. The total is 4 but the makeup is 2 B, 2 not blue. This is wrong, because the odds of two blue are 1 to 5, not 50-50 (or 1 to 1 in lowest terms)

No makeup of just 2 blue results in 50-50 odds, so I thought it might be impossible. Then I thought "nPr", which is the symbol on a TI-83 for a permutation of r items out of a total of n, where you pick and arrange the r items. I then realized this involved not a permutation, but a combination (where order does not matter), or nCr on a TI-83 where the number of blue was more than 2.

Using 3 blue, this creates 3 possible ways to pick 2--if you numbered them 1, 2, and 3, they would be 1 and 2, 1 and 3, and 2 and 3.

To have 50-50 odds, you need 3 that don't contain 2 blues (not necessarily NO blues, just at least one that isn't blue). Eureka! Just add one non blue (call it 4), and paired with each of the 3 blues, you have 3 pairs that don't have 2 blues--1 and 4, 2 and 4, and 3 and 4. Putting it all together, the number of pairs with 2 blues (3) and the number of pairs that have fewer than 2 blues (3) results in 50-50 odds. In mathematical symbols, this is C(3,2)/C(3,2) (what you would see on EXCEL). The probability of getting a pair of blues is .5, obtained from C(3,2)/C(4,2).

The FACT of the matter is that all permutation and combination calculations are defined in terms of FACTORIALS (n factorial--shown in symbols as n! = the product of the first n integers). For example, 5! = 5*4*3*2*1 = 120, 6! = 6*5*4*3*2*1 = 720, etc.. However, the numbers in the solution are small enough that you can use simple multiplication without using factorials.

This being World Series week presents an interesting factorial challenge that for you that involves numbers and formulas but doesn't involve baseball or calculations. The formula for the number of combinations of r items out of n is C(n,r) = n!/(r!*(n-r)!). It can be shown algebraically that C(n,r) = C(n, n-r). For example, C(6,2) = C(6,4) = 15.

The winner of the World Series is the first team to win 4 games out of 7 (same as the NBA and Stanley Cup finals). From the perspective of the winning team, there are C(7,4) = 35 ways that this can be done. Per the previous paragraph C(7,4) = C(7,3). Your challenge: show why this is true without listing which games the winner won or doing any calculations--just logic. It's actually very simple. And yes, I am a math professor and teach this topic--in fact it's one of my favorite topics. However, when I ask my students why this is so sans algebra or numbers, using only logic, they're unable to do so. But all of you are math savvy enough that you can do it!

Bonus: the World Series ends when a team wins 4 games. Thus, it can be over in 4, 5, 6, or 7 games. How many ways can each scenario result? Your hint: There is only one way for the series to end in 4 games and that is if the champion wins all of the first 4 games. That leaves 34 other possibilities if it goes 5, 6, or 7 games. Clarification: all scenarios are from the perspective of the winning team. Thus, there are technically 2 ways it can be over in 4 games using this year's teams--if the Giants win the first 4 or the Royals win the first 4. Consider only one team when answering this challenge.

If a team wins 4 games, it does not win 3 games, making C(7,4) = C(7,3). More generally, C(n,r) means picking r objects from n, leaving n-r objects not picked. If you reversed roles, you'd be picking n-4 objects and leaving n objects not picked--thus (C(n,r)=C(n,n-r). No numbers or algebra was used in this explanation, nor were any animals harmed in doing so.

The World Series question: to win the Series in 5 games, your first inclination is C(5,4)=5. However, that includes the possibility of a 4 game sweep. The key to the answer is that to win the Series in 5 games, the last game won is the fifth game, with 3 wins (or equivalently, 1 loss) in the first 4 games. This equals C(4,3) = C4,1) = 4.

The same logic applies to win the Series in 6 games and 7 games--3 victories in the first 5 games or 6 games, respectively (or equivalently, 2 or 3 losses in the first 5 or 6 games, respectively). In symbols, the numbers are C(5,3) = C(5,2)=10 and C(6,3) = 20, respectively, Thus, the number of ways of winning the Series in 4, 5, 6, and 7 games is 1, 4, 10, and 20, respectively. These translate to probabilities of 1/35, 4/35, 10/35, and 20/35, respectively.

In 2003, a Harvard Math Professor (whose name was never revealed for obvious reasons--read on) committed a worse error than Bill Buckner's famous 1986 error, by stating that these probabilities were 2/16. 4/16, 5/16, and 5/16. I can only wonder with results like this, if he so-called math Professor obtained his Ph.D. from one of those bogus diploma mills that grants degrees to everyone who forks over the $, no questions asked.

Allergan makes an eye drop called Lumigan, to treat glaucoma. They noticed that a side effect of the medication is thickening of eyelashes. So now, they market the same drug as Latisse, to thicken your eyelashes, at $114 per 3 ml dropper, without the need for expensive mascara. Both of these have the side effect of turning blue eyes brown, so if that ever becomes a thing, I'm sure they'll find another name for the med, and market it for that.

I have no doubt that someone (though not me) can solve this puzzle mathematically. However, mathematics is not necessary to solve it. At root, this is a logical puzzle and only logic is needed to solve it.

It may well be fun to find more than one answer but it does not seem to be a requirement of the puzzle. A single answer would appear to satisfactorily answer the question, “How many students are in the class?”

Over pizza last night we proved the number that we came up with by placing equal and sweet and low packets in a hat. Equal represented blue eyes and sweet and low, pink eye. Double blue was one ahead until we shook the hat so vigorously that one of each popped out. Truly random as far as we are concerned.

It seems a bit ironic to me that the Underground Railroad was as successful as it was well over a hundred years ago, but here in Seattle in the twenty-first century we are still unable to free Big Bertha.

Good stuff above. I think the puzzle could have been better worded and constrained. In fact, I think it would have been a better "Car Talk" puzzle, perhaps something like this:

Ray: Let’s say you have a regular grade school classroom, and two students are chosen at random.Tom: I hope they’re not in trouble! They’re not brothers, are they?Ray: No. Well, maybe. But anyway, a regular grade school classroom…so, even with budget cuts, let’s say there’s less than 30 students in the room.Tom: Ok.Ray: Those 2 students both happen to have blue eyes. And, as it turns out, the odds are exactly 50:50 that 2 students chosen at random will have blue eyes. Tom: Ok. And otherwise, the students chosen…Ray: They might both have brown eyes; there might be one set of blue-eyes, one set of brown; they might…Tom: They might both have pink eye!Ray: Sure. Maybe they all do! So anyway, the question is, how many students are in the class?

Very clever, dear WW. So Enya and Weird Al-fan got a nice shout-out from Will on their "elegant" answer: Romania, etc. in last week's puzzle. I'm just gonna take a stab at this one. Once dated a girl with different-colored eyes. Sigh...the Ox that got away.

Since you delete your below comment I figured you had finally understood your error in thinking, but now I see you are still confused and in the dark. I have a suggestion for you. Substitute the answer you now have, if you even have one, for one you would get had the puzzle been presented in the form of the kids in the classroom wearing blue shirts and some other color shirts. If you come up with the same answer, then you are most likely correct, but if not then you are wrong.

There is currently a crisis among the Canadian ice farmers, due to global warming. This is little known outside of Canada. Block and cubed ice is a major Canadian export, but the growing and harvesting season has been shrinking. The farmers have resorted to decreasing the size of block ice from 10 pounds to 4 kilograms, allegedly to change to the metric system, but in fact a scheme to reduce the standard size by 12%. Late in the season, they stop growing block ice altogether, which has a greater yield per acre than cubed ice and is easier to harvest. All this is happening just as the craft cocktail scene is expanding in the United States, increasing demand.

Reminds me of the time my very young niece was playing hide-and-seek with my tall son. He hid in the front hall closet. She opened the closet door, revealing his legs, though his head and torso were hidden behind the coats. She appeared stumped. My wife tried to help. "Do you think he's in the closet?" "Maybe half," my niece replied.

There are FOUR students in the class. THREE of them have BLUE EYES. One student’s eyes are some other UNKNOWN COLOR.

There are three possibilities of UNKNOWN COLOR/BLUE. There are three possibilities of BLUE/BLUE.

The Breakdown:

UC with B1 \UC with B2 > = 3 or 50%UC with B3 /

B1 with B2 \B1 with B3 > = 3 or 50%B2 with B3 /

There are no other combinations possible, therefore we have liftoff.

I cannot help wondering how many submitted FOUR as their answer without any further explanation as to how they arrived at their conclusion and they actually believe there are 2 BLUE & 2 UNKNOWN COLOR students in the classroom. Of course this would not be correct.

Anyone who became caught up in attempting to determine the answer by considering the statistics regarding having a child with blue, brown, grey, hazel etc. colored eyes was falling into a trap similar to a red herring. This trap may, or may not, have been intentional, but I would bet several were caught up in it even though it was only vaguely implied and not stated.

This puzzle did not need to be tied to eye color. It could just as well be about coins, or marbles, or perhaps black and orange jelly beans in a hollowed out pumpkin. The same logic applies. It might be well to keep in mind that Will Shortz did not present us with a Martin Gardiner puzzle, but he gave us a puzzle “based on a puzzle from a Martin Gardner book.” He somehow changed the original puzzle and it most likely was more elegant in its original presentation. Although I see this puzzle as being flawed in its presentation I enjoyed it just the same.

A SMALL CLASS OF 4 STUDENTS: 3 blue-eyed students and 1 non-blue-eyed student. If you select two students, then there are 6 possibilities: B1+B2, B1+B3, B2+B3, B1+H (Hazel-eyed), B2+H, B3+H. So 3 out of 6 possibilities yields two blue-eyed students, and 3 out of 6 possibilities yields a blue-eyed student and a hazel-eyed student. Thus the chances are 50-50 that a pair of blue-eyed students will be selected at random from this class of 4 students.

I think jan and I have already given the ultimate hints; but I'll offer one more:

Just as there are infinitely many numbers which are squares, but are also triangular numbers (i.e. 1, 36, 1225, etc.), and just as there are infinitely many Pythagorean triples in which the first two numbers differ by just one (i.e. 3, 4, and 5; 20, 21, and 29; 119, 120, and 169; etc.), so there are also infinitely many class-sizes with associated lesser numbers of blue-eyed students within such classes so that randomly picking two of them gives exactly even odds of both having blue eyes.

Before I reveal my submission to the NPR website, I wanted to first explain my last hint. It was more than just two examples of series which go on forever; it was meant to lead you, the reader to yet ANOTHER series (actually TWO series) which are quite useful in finding the terms of those other two series.

The first example I gave, numbers which are squares, but are also triangular numbers (i.e. 0, 1, 36, 1225, etc.), should've led you to 0*1, 1*1, 4*9, 25*49, ... which could lead you to

─ Numerators: 1 1 3 7 ...Denominators: 0 1 2 5 ...

Two series of numbers in which the square of the top one is twice the square of the bottom one +/- 1.

The next series would lead you there also!

What you see above, which I have labeled numerators and denominators, form a series of fractions (except for the 1 and 0 beginning the series) which approach the square root of 2.

Anyway, the name of those two series: What I've called the denominators is called The Pell Numbers, and what I've called the numerators are their Half Companions.

What follows is exactly what I submitted to the NPR site, after first quoting the puzzle. The only difference was after my submission I noticed that I had omitted a 1 between a + and a ). Other than correcting that, what you see below is exactly what I submitted.

While some may answer 4, of which 3 have blue eyes, while that is certainly not the answer you're expecting, I nevertheless think you should accept that as an alternate answer - after all, a size-4 class COULD happen!

I'm sure the expected answer is 21, of which 15 have blue eyes. In fact, if there is no limit to the class size, then there are INFINITELY MANY solutions!

They can be generated by consecutive pairs of Pell numbers with their Half-companions.

n 0 1 2 3 4 5 6 ...

H(n) 1 1 3 7 17 41 99 ...P(n) 0 1 2 5 12 29 70 ...

Ignoring the 0 terms, the cross-products { P(n)*H(n+1) and H(n)*P(n+1) } are two consecutive numbers; while the parallel-products { H(n)*H(n+1) and 2*P(n)*P(n+1) } are two consecutive numbers whose product with each other is obviously twice the product of the first consecutive numbers.

So the 1st solution is a class-size of 4, of which 3 have blue eyes.

Probability = 3/4 * 2/3 = 1/2.

The 2nd solution is what I'm sure is THE EXPECTED SOLUTION: a class-size of 21, of which 15 have blue eyes.

Brilliant, thanks! I graphed the solution and noticed the sqrt(2) slope (also noted in the hints here, and obvious from taking the limit), but was unable to find the integer points. Thanks for the education on Pell numbers! Very elegant.

I think it's nearly certain that the answer of "4" will not be accepted, on the grounds that 4 students is not a "regular" grade school class, as the problem specifies. 21 will presumably be the only answer deemed correct.

I'm predicting a small number of correct answers, if that's their decision--maybe around 200-300.

I agree with 15/21 being the best answer. I do think the puzzle could have been worded better. It's uncommon but not impossible to have classrooms of 4, and it wouldn't be unreasonable for a person to stop trying to solve the puzzle once they reached that conclusion. So an extra puzzle detail that I added in my Car Talk humour attempt would have been additionally suggesting that there was a finite probability of choosing 2 non-blue eyed students, which would have more subtly implied a classroom of more than 4 students. An upper bound would have been useful as well but I think most people would question a class size of 120.

3 does not even come close to working. There would be only one chance of coming up with Blue/Blue, but two chances of Brown/Blue.

If our winner this week is correct and WS does not even mention 4 as being the answer, then he has really slipped even further in my disrespect for his sloppy work. It now appears he was presenting us with more of a mathematical puzzle than a logic puzzle, which is what I, and apparently most others, assumed it to be. I suspect he again, as usual, did not check his possibilities and just assumed the answer would be the one he has so arrogantly chosen over all other equally correct answers. This guy is really an egotistical jerk as I have been informed is the case from a third party who should know.

I was a bit surprised last Thursday when you placed your answer of 4 at the beginning. I'm even more surprised now that your answer of 4 remains at the top of this blog long after even when many of us have identified a class size of 21 as the expected answer; especially since this week's chosen winner, Jake, says of Will that "He doesn't accept 4 kids".

Both answers are perfectly correct. If you prefer a mathematical puzzle, then you may prefer the answer of 21, but if you are more inclined toward using logic as your means of discerning an answer, then 4 is the obvious answer. I prefer the latter as I am not schooled at math, but I enjoy solving problems using my intellect. Were I proficient at math I would still prefer 4 as the better answer. I cannot speak for Blaine, but I do know he enjoys math puzzles, however he is obviously a logical thinker and has no reason to back away from his pristine answer of 4. To my way of thinking it is far more satisfying to lie in bed, as I did, and solve this puzzle with nothing but my reasoning instead of resorting to a calculator to do the work. Insisting that only one of these answers can be the correct one is, in my opinion, both arrogant and obnoxious on the part of Mr. Shortz.

I think I should also add that if WS was going to insist on the answer of 21 only, then he should have stated the classroom was filled with students. Many puzzles imply something such as a room being filled, but the answer turns out to be much different, requiring the solver to think outside the proverbial box, as it were. WS only stated the classroom was typical, but gave no factual evidence to indicate anything more regarding its content such as the size of the class itself, or the percentage of students present.

If the classroom has 15 out of 21 students with blue eyes, it is already not typical. The problem could have asked about pairs with brown eyes, rather than blue, for it to have had a chance to be a typical classroom.

Next week's challenge: This challenge comes from listener Mike Reiss, who's a writer for The Simpsons. Name a well-known TV actress of the past. Put an R between her first and last names. Then read the result backward. The result will be an order Dr. Frankenstein might give to Igor. Who is the actress, and what is the order?